PRESENTED BY:

ANZAR SHAH

41 IX-C

CIRCLES

Circles And Its Related Terms.

1.1 CIRCLE:

The collection of all the points in plane, which are at a fixed distance from a fixed point in the plane, is called a circle.

1.2 Radius, Diameter, Center, Chord, Arc And Sector.

The line segment joining the centre and any point on the circle is also called a radius of the circle.

Diameter is two times the radius.

The fixed point is called the centre of the circle.

If we take two points P and Q on a circle, then the line segment PQ is called a chord of the circle.

A piece of a circle between two points is called an arc.

The region between an arc and the two radii, joining the centre to the end points of the arc is called a sector.

THEOREM 1

To Prove: Equal chords of a circle subtend equal angles at the centre.

Proof : We are given two equal chords AB and CD of a circle with centre O. We want to prove that:

∠ AOB = ∠ COD. In triangles AOB and COD,

OA = OC (Radii of a circle) OB = OD (Radii of a circle) AB = CD (Given) Therefore, Δ AOB Δ COD ≅ (SSS rule) This gives ∠ AOB = ∠ COD (CPCT)

THEOREM 2

If the angles subtended by the chords of a circle at the centre are equal, then the chords are equal.

Theorem 2 is converse of the Theorem 1.

THEOREM 3TO PROVE: The line drawn through the centre of a circle to

bisect a chord is perpendicular to the chord.PROOF: Let AB be a chord of a circle with centre O and O is

joined to the mid-point M of AB. We have to prove that OM ⊥ AB. Join OA and OB. In triangles OAM and OBM, OA = OB (Given) AM = BM (Given) OM = OM (Common) Therefore, ΔOAM ≅ ΔOBM (SSS) This gives ∠OMA = ∠OMB = 90° (CPCT)

THEOREM 4 TO PROVE: Equal chords of a circle (or of congruent

circles) are equidistant from the centre (or centers).

PROOF: Take a circle of any radius. Draw two equal chords AB and CD of it and also the perpendiculars OM and ON on them from the centre O.

Divide the figure into two so that D falls on B and C falls on A. We observe that O lies on the crease and N falls on M. Therefore, OM = ON.

THEOREM 5Chords equidistant from the centre of a circle are equal in length.

Theorem 5 is converse of the Theorem 4

The angle subtended by an arc at the centre is defined to be angle subtended by the corresponding chord at the centre in the sense that the minor arc subtends the angle and the major arc subtends the reflex angle.

The angle subtended by the minor arc PQ at O is ∠POQ and the angle subtended by the major arc PQ at O is reflex angle POQ.

We conclude that congruent arcs (or equal arcs) of a circle subtend equal angles at the centre.

THEOREM 6TO PROVE: The angle subtended by an arc at the centre is double the angle

subtended by it at any point on the remaining part of the circle.PROOF: Given an arc PQ of a circle subtending angles POQ at the centre O and PAQ at a

point A on the remaining part of the circle. Consider the three different cases arc PQ is minor(Fig 1), arc PQ is a semicircle(Fig 2) and in arc PQ is major(Fig 3). Let us begin by joining AO and extending it to a point B.

In all the cases, ∠ BOQ = ∠ OAQ + ∠ AQO because an exterior angle of a triangle is equal to the sum of the two interior opposite angles. In Δ OAQ, OA = OQ (Radii of a circle) Therefore, ∠ OAQ = ∠ OQA This gives ∠ BOQ = 2 ∠ OAQ (1) Similarly, ∠ BOP = 2 ∠ OAP (2) From (1) and (2), ∠ BOP + ∠ BOQ = 2(∠ OAP + ∠ OAQ) This is the same as ∠ POQ = 2 ∠ PAQ (3) For the case (iii), where PQ is the major arc, (3) is replaced by reflex angle POQ = 2 ∠ PAQ.

THEOREM 7TO PROVE: Angles in the same segment of a circle are equal.PROOF: According to Theorem 6, ∠ PAQ = 1/2 ∠POQ = 1/2×180°=90°.

If we take any other point C on the semicircle, again we get that ∠PCQ = 90°(Fig 1).Suppose we join points P and Q and form a chord PQ in the figures used in Theorem 6. Then ∠PAQ is also called the angle formed in the segment PAQP.A can be any point on the remaining part of the circle. So if we take any other point C on the remaining part of the circle, ∠POQ = 2 ∠PCQ = 2 ∠PAQTherefore, ∠PCQ = ∠PAQ.

THEOREM 8If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, the four points lie on a circle (i.e. they are concyclic).

Theorem 8 is converse of the Theorem 7

THE END